Question

Which is one solution of the following system 2y-2x=12 x^2+y^2+36

Answer 1

Answer:

Given the system of equations:

$2y-2x =12$              .....[1]

$x^2+y^2=36$          .....[2]

We can write equation [1] as;

2(y-x) = 12

Divide both sides by 2 we get;

$y-x = 6$    

or

y = x+6                                        ....[3]

Substitute equation [3] in [2] we get;

$x^2+(x+6)^2=36$

$x^2+x^2+36+12x=36$

Subtract 36 from both sides we have;

$x^2+x^2+12x=0$

Combine like terms;

$2x^2+12x =0$

or

$2x(x+6)=0$

By zero product property:

x =0 and x+6 = 0

or

x = 0 and x = -6

now, substitute the given values of x in [3] we have;

for x = 0

y = 0+6 = 6

for x = -6

y = -6 + 6 = 0

Therefore, the solution for the given system of equation is, either (0, 6)  or (-6, 0)